Let tangents drawn from origin to circle, x^{2}+y^{2}-8 x-4 y+16=0 touch it at points A and B . (A B

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10-1.Circle and System of Circles

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Let the tangents drawn from the origin to the circle, $x^{2}+y^{2}-8 x-4 y+16=0$ touch it at the points $A$ and $B .$ The $(A B)^{2}$ is equal to

A

$\frac{52}{5}$

B

$\frac{32}{5}$

C

$\frac{56}{5}$

D

$\frac{64}{5}$

(JEE MAIN-2020)

Solution

$\mathrm{R}=\sqrt{16+4-16}=2$

$\mathrm{L}=\sqrt{\mathrm{S}_{1}}=4$

$\mathrm{AB}(\text { Chord of contact })=\frac{2 \mathrm{LR}}{\sqrt{\mathrm{L}^{2}+\mathrm{R}^{2}}}=\frac{8}{\sqrt{5}}$

$(\mathrm{AB})^{2}=\frac{64}{5}$

Standard 11

Mathematics

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